Edit: With apologies for totally rewriting the question:
Suppose $f_n$ is a bounded sequence in $L^1(\omega)$, where $\omega$ is an open subset of $\mathbb R^d$. Need $(f_n)$ have an almost-everywhere-convergent subsequence?
(Of course if $L^1$ were reflexive we could start by applying Banach-Alaoglu to get a weak-* convergent subsequence. But it's not so we can't.)
Let $\omega$ be an open of $R^n$ and let $f_n$ be a bounded sequence in $L^1(\omega)$, As $L^1(\omega)$ is not refexive we can not ecxtract a convergente seubsequence, how can I deal with that. Thanks.
No. Counterexample in $L^1([0,1])$: Consider the Rademacher functions:
$$r_n=\sum_{j=0}^{2^n-1}(-1)^j\chi_{I_{n,j}},$$where $$I_{n,j}=[j2^{-n},(j+1)2^{-n}).$$If $r_{n_j}\to f$ almost everywhere then dominated convergence implies that $||r_{n_j}-f||_1\to0$, which is impossible since $$||r_n-r_m||_1=1\quad(n\ne m).$$
(In fact with a little more work one can show that any subsequence is almost everywhere divergent. The little more work is in showing that $$\lim_{n,m\to\infty, n\ne m}\int_E|r_n-r_m|=|E|,$$where $|E|$ is the measure of $E$. Edit: Just realized this is much more obvious than I thought. Note that $|r_n-r_m|=\frac12(r_n-r_m)^2=1-r_nr_m$ and recall that finite products of Rademacher functions are orthogonal (these are the Walsh functions.))
Note: In fact we do have $r_n\to0$ weakly in $L^1$. So it seems that it wouldn't help much if $L^1$ were reflexive; convergence in the typical weak or weak-* topology evidently has very little to do with convergence almost everywhere.